\(2\dfrac{1}{3}.\left(x+\dfrac{1}{2}\right)x=3\dfrac{3}{4}\)

<=> \(\dfrac{7}{3}x\left(x+\dfrac{1}{2}\right)=\dfrac{15}{4}\)

<=> \(\dfrac{7}{3}x^2+\dfrac{7}{6}=\dfrac{15}{4}\)

<=> \(\dfrac{7}{3}x^2=\dfrac{15}{4}-\dfrac{7}{6}\)

<=> \(\dfrac{7}{3}x^2=\dfrac{31}{12}\)

<=> x² = \(\dfrac{31}{12}:\dfrac{7}{3}\)

<=> x² = \(\dfrac{31}{28}\)

<=> x = \(\dfrac{\sqrt{217}}{14}\)

Hỗn số hết nha

\(5\dfrac{1}{3}+3\dfrac{7}{17}-1\dfrac{7}{17}+4\dfrac{2}{3}\)

= \(\dfrac{16}{3}+\dfrac{58}{17}-\dfrac{24}{17}+\dfrac{14}{3}\)

= \(\dfrac{16}{3}+\dfrac{14}{3}+\dfrac{58}{17}-\dfrac{24}{17}\)

= \(\dfrac{30}{3}+\dfrac{34}{17}\)

= 10 + 2

= 12

a. (0,125)² . 64

= \(\dfrac{1}{64}.\dfrac{64}{1}\)

= \(\dfrac{1.1}{1.1}=1\) 

cần điều kiện nha

Điều kiện đâu

\(\sqrt{x^2+8x+16}=12\)     ĐKXĐ: x \(\ge-4\)

<=> \(\sqrt{\left(x+4\right)^2}=12\)

<=> x + 4 = 12

<=> x = 12 - 4

<=> x = 8

Bn vào thiết lập nha.